Question

The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism: where \(\mathrm{A}=\) adenosine. If the \(\mathrm{p} K_{a}\) for this anion is \(7.21\). what is the ratio of \(\left[\mathrm{AMP}-\mathrm{OH}^{-}\right]\) to \(\left[\mathrm{AMP}-\mathrm{O}^{2-}\right]\) in blood at \(\mathrm{pH} 7.4\) ?

Short answer

The ratio of \([\mathrm{AMP}-\mathrm{OH}^{-}]\) to \([\mathrm{AMP}-\mathrm{O}^{2-}]\) in blood at pH 7.4 is approximately 1.54, meaning there are about 1.54 times more \(\mathrm{AMP}-\mathrm{OH}^{-}\) ions than \(\mathrm{AMP}-\mathrm{O}^{2-}\) ions at this pH level.

Step-by-step solution

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given as: \[pH = pK_a + \log \frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\] In this case, the base is \([\mathrm{AMP}-\mathrm{O}^{2-}]\) and the acid is \([\mathrm{AMP}-\mathrm{OH}^{-}]\).

Plug in the known values into the equation

We've been given the \(pK_a\) value of the anion (7.21) and the pH of the blood (7.4). Substitute these values into the Henderson-Hasselbalch equation: \[7.4 = 7.21 + \log \frac{[\mathrm{AMP}-\mathrm{O}^{2-}]}{[\mathrm{AMP}-\mathrm{OH}^{-}]}\]

Solve for the ratio

Subtract 7.21 from both sides of the equation: \(0.19 = \log \frac{[\mathrm{AMP}-\mathrm{O}^{2-}]}{[\mathrm{AMP}-\mathrm{OH}^{-}]}\) Next, we need to find the antilog (inverse log) of both sides to solve for the ratio: \[\frac{[\mathrm{AMP}-\mathrm{O}^{2-}]}{[\mathrm{AMP}-\mathrm{OH}^{-}]} = 10^{0.19}\] Calculate the value of \(10^{0.19}\): \[\frac{[\mathrm{AMP}-\mathrm{O}^{2-}]}{[\mathrm{AMP}-\mathrm{OH}^{-}]} \approx 1.54\]

Interpret the result

The ratio of \(\left[\mathrm{AMP}-\mathrm{OH}^{-}\right]\) to \(\left[\mathrm{AMP}-\mathrm{O}^{2-}\right]\) in blood at pH 7.4 is approximately 1.54. This implies that there are about 1.54 times more \(\mathrm{AMP}-\mathrm{OH}^{-}\) ions than \(\mathrm{AMP}-\mathrm{O}^{2-}\) ions in the blood at this pH level.

Key concepts

pKa and pH relationship

The pKa and pH relationship is fundamental in chemistry, especially when dealing with acid-base equilibrium and buffer solutions. The pKa of a substance is a constant that provides insight into the strength of an acid. It is the pH at which half of the acid molecules are dissociated. This means half of the molecules donate a proton to become the conjugate base. A lower pKa indicates a stronger acid.
In the context of the Henderson-Hasselbalch equation, pKa is used to determine how the ratio of an acid-base pair will change with pH. The equation is: \[ pH = pK_a + \log \frac{[\text{Base}]}{[\text{Acid}]} \] This equation helps in calculating the ratio between the concentration of the acid and its conjugate base. When the pH equals the pKa, the concentrations of the acid and base are equal. Thus,

• If pH < pKa, the solution is acidic and more of the acid form is present.
• If pH > pKa, the solution is basic and more of the base form is present.

For example, if discussing AMP ions as in the exercise, knowing the pH relative to the pKa will tell us how the system behavior changes with pH adjustments in metabolic processes.

phosphate metabolism

Phosphate metabolism is critical for maintaining cellular functions since phosphates are key components of many biological molecules such as nucleotides like AMP. During metabolism,

• Phosphates act as energy carriers in the forms like ATP and AMP.
• They are also crucial in signal transduction processes that control cellular activities.

The metabolism of phosphate involves complex biochemical processes where phosphate ions can switch between different oxidation states, affecting their chemical reactivity. Phosphate metabolism is regulated by hormones like parathyroid hormone and vitamin D which control phosphate absorption and excretion.
In terms of acid-base balance, phosphates can act as buffers. They can accept and donate protons to maintain stable pH levels in biological fluids like blood. This buffering capacity is key to homeostasis, preventing drastic pH changes that could disrupt metabolic reactions.

acid-base equilibrium

Acid-base equilibrium is crucial in chemistry, particularly within biological systems where enzymes and biochemical processes rely on specific pH ranges to function optimally.
Here's how it works:

• Acids, known as proton donors, release hydrogen ions (H⁺) into solutions.
• Bases, on the other hand, are proton acceptors and can neutralize acids by removing H⁺.

The equilibrium between acids and bases is illustrated not just by the concentrations but by their ability to resist changes in pH when an acid or base is added—a property known as buffering capacity. The Henderson-Hasselbalch Equation outlined earlier is pivotal in understanding these interactions by relating pH, pKa, and the ratio of acid to base. This equation is particularly useful for designing buffer solutions with desired pH levels.
Within the body, maintaining acid-base equilibrium is vital. The kidneys and lungs play significant roles by removing excess acids or bases to keep blood pH within the narrow range necessary for healthy functioning. In the given exercise, the ratio determined using this equation reflects the specific acid-base balance involving phosphate ions in the blood, showcasing the natural buffering and regulatory mechanisms in action.